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Friday, 9 October

JEE Main 2021 · PhysicsMultiple choiceSingle correctMediumCalculation

JEE Main 18 March 2021, Shift 1, Physics Q9

Question 9 of 90 in this shift, Physics question 9 of 30, Section A.

The time period of a simple pendulum is given by T=2πlgT = 2\pi\sqrt{\frac{l}{g}}. The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is,
  1. (1)2%
  2. (2)3%Official answer
  3. (3)4%
  4. (4)5%

Official answer

Option 2

NTA final key.

Chapter
Units and Measurements
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.